Is my proof for boundedness correct?

Question

I was given this question:

Let $h \in L^{1}([0,1], m)$ and let $F$ be a linear functional on the normed space $C[0,1]$ (with the maximum norm), defined by $F(g) = \int_{0}^{1} hg.$ Prove that $F$ is a bounded linear functional on $C[0,1]$ and determine its norm.

My trial:

Since $|F(g)| = |\int_{0}^{1} h(t)g(t) dt| \leq \int_{0}^{1}|h(t)||g(t)| \leq \max_{t\in [0,1]} |g(t)|.\int_{0}^{1} |h(t)|(1-0) = \||g\|_{max} \|h\|.$

Therefore, $\|F\| \leq \|h\|.$

Answer 1

Everything you've done so far is correct. To finish, assume first $0=x_0<x_1<...<x_n=1$ such that $h=\sum_{k=1}^n c_k1_{[x_{k-1},x_k)}$ for some constants $c_k$. Let $g_n$ be some continuous function such that $g_n\equiv \frac{c_k}{|c_k|}$ on $[\frac{n+1}{n}x_{k-1}, \frac{n-1}{n}x_k]$ for every $k$ such that $c_k\neq 0$ and $0$ otherwise. Furthermore, we impose that $|g_n|\leq 1$. Such a function can be constructed as piecewise affine.

Then, $\|g_n\|_{\infty}=1$ for all $n$ and $\int h g_n\textrm{d}m\to \int |h|\textrm{d}m$ implying the desired in this case.9

Now, functions of this form are dense in $L^1([0,1],m)$, since the half-open intervals generate the Borel $\sigma$-algebra and the measure is finite and the constant $1$ function is almost everywhere equal to $1_{[0,1)}$. Note, furthermore, that if $h_n\xrightarrow{L^1} h$, we have, for all continuous $f$ that

$$ \left| \int h_nf \textrm{d}m-\int h f\textrm{d}m\right|\leq \|h_n-h\|_{L^1} \|f\|_{\infty} $$ This implies that the map $\varphi(h)= F_h$ is a continuous map from $L^1$ to $C([0,1])^*$. Since $\varphi$ is an isometry on the dense set of simple functions of the above form, then $\varphi$ is an isometry. This concludes the proof.

Answer 2

It is not too hard to show that if $\eta$ is continuous then $\|F\| = \|\eta\|_1$ using a sequence of approximations to $\operatorname{sgn} \eta$ (or the complex equivalent).

The continuous functions are dense in $L^1[0,1]$, so there is a sequence of continuous functions $h_n$ such that $\|h-h_n\|_1 \to 0$.

Let $\epsilon>0$ and choose a continuous $h_n$ such that $\|h-h_n\|_1 < \epsilon$ and so $\|h_n\|_1 > \|h\|_1 -\epsilon$. Now choose $g$ with $\|g\|_\infty = 1$ such that $| \int h_n g | \ge \|h\|_1-\epsilon$.

We have $||\int h g| - |\int h_n g|| \le |\int h g - \int h_n g| \le \|g\|_\infty \|h-h_n\|_1 < \epsilon $.

Combining, $|\int hg| \ge |\int h_n g|-\epsilon \ge \|h\|_1 - 2 \epsilon$ and so $\sup_{\|g\|_\infty \le 1} |\int hg| = \|h\|_1$.